You have found the following ages (in years) of all 6 seals at your local zoo: $ 6,\enspace 4,\enspace 12,\enspace 1,\enspace 23,\enspace 3$ What is the average age of the seals at your zoo? What is the variance? You may round your answers to the nearest tenth.
Because we have data for all 6 seals at the zoo, we are able to calculate the population mean $({\mu})$ and population variance $({\sigma^2})$ To find the population mean , add up the values of all $6$ ages and divide by $6$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6}} $ $ {\mu} = \dfrac{6 + 4 + 12 + 1 + 23 + 3}{{6}} = {8.2\text{ years old}} $ Find the squared deviations from the mean for each seal. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $6$ years $-2.2$ years $4.84$ years $^2$ $4$ years $-4.2$ years $17.64$ years $^2$ $12$ years $3.8$ years $14.44$ years $^2$ $1$ year $-7.2$ years $51.84$ years $^2$ $23$ years $14.8$ years $219.04$ years $^2$ $3$ years $-5.2$ years $27.04$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{4.84} + {17.64} + {14.44} + {51.84} + {219.04} + {27.04}} {{6}} $ $ {\sigma^2} = \dfrac{{334.84}}{{6}} = {55.81\text{ years}^2} $ The average seal at the zoo is 8.2 years old. The population variance is 55.81 years $^2$.